If A + B + C = 180° prove that
Sin2A + sin2B – sin2C = 4cosAcosBsinC
4cosAcosBsinC
To prove the equation Sin2A + sin2B - sin2C = 4cosAcosBsinC, we will start with the given equation A + B + C = 180° and use trigonometric identities to manipulate the equation until we get the desired result.
Step 1: Expressing sin2A, sin2B, sin2C in terms of cosA and cosB
We know that sin2θ = 2sinθcosθ, so we can express sin2A, sin2B, and sin2C in terms of cosA and cosB:
sin2A = 2sinAcosA,
sin2B = 2sinBcosB,
sin2C = 2sinCcosC.
Step 2: Substituting sin2A, sin2B, sin2C in the equation
Substituting the expressions for sin2A, sin2B, and sin2C in the equation, we have:
2sinAcosA + 2sinBcosB - 2sinCcosC = 4cosAcosBsinC.
Step 3: Simplifying the equation
Next, we can factor out a 2 from the left-hand side of the equation:
2(sinAcosA + sinBcosB - sinCcosC) = 4cosAcosBsinC.
Step 4: Using the sum-to-product identities
Now, let's focus on the expression inside the parentheses, sinAcosA + sinBcosB - sinCcosC. To simplify this expression, we will use the sum-to-product identities:
sin α cos β = (1/2)(sin(α + β) + sin(α - β))
Applying this identity, our expression can be rewritten as:
sinAcosA + sinBcosB - sinCcosC = (1/2)(sin(A + A) + sin(A - A) + sin(B + B) + sin(B - B) - sin(C + C) - sin(C - C))
Since sin(2θ) = 2sinθcosθ, we can simplify the equation further:
sinAcosA + sinBcosB - sinCcosC = (1/2)(2sin2A + 2sin2B - 2sin2C).
Step 5: Substituting the simplified expression back into the equation
Substituting the simplified expression back into the equation, we get:
2(sin2A + sin2B - sin2C) = 4cosAcosBsinC.
Step 6: Dividing both sides of the equation by 2
Finally, we divide both sides of the equation by 2 to isolate the desired expression:
sin2A + sin2B - sin2C = 2cosAcosBsinC.
Therefore, we have proved that Sin2A + sin2B - sin2C = 2cosAcosBsinC using the given equation A + B + C = 180° and trigonometric identities.